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I am trying to find all the eigenvalues of $P$ defined below:

$$P=\begin{bmatrix} 0.5&0.5&0&0&\cdots 0&0 \\ 0.25&0.5&0.25&0&\cdots 0&0\\ 0&0.25&0.5&0.25&\cdots 0&0\\ \vdots \\ 0&0&0&0&\cdots 0.5&0.5 \end{bmatrix}_{n\times n}$$

So $P$ has $0.5$ along the main diagonal. It has $0.25$ on diagonals above and below the main diagonal except for the first and last row. Hence $P$ is not exactly a Toeplitz matrix.

My attempt: A paper I'm looking at, gives eigenvalues of the following Toeplitz matrix

$$ Q=\begin{bmatrix} b&a \\ c&b&a \\ &\ddots&\ddots&\ddots \\ &&c&b&a \\ &&&c&b \end{bmatrix} $$

as $\lambda_j = b+2\sqrt{ca}\cos\left(\frac{j\pi}{n+1} \right)$

So I'm wondering if I will be able to find eigenvalues of $P$ even though its not a Toeplitz matrix precisely?

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  • $\begingroup$ I suspect that this wikipedia page could be helpful $\endgroup$ – Ben Grossmann May 11 at 14:25
  • $\begingroup$ You have a typo in the eiegenvalues of the Toeplitz matrix, the coefft of the $\cos$ has to be $2\sqrt{ac}$. $\endgroup$ – ancient mathematician May 11 at 14:26
  • $\begingroup$ There's clearly an eigenvalue $0$ with multiplicity $1$ and another $1$ of multiplicity $1$. $\endgroup$ – ancient mathematician May 11 at 14:28
  • $\begingroup$ @ancientmathematician Thanks, corrected the typo. $\endgroup$ – manifolded May 11 at 14:31
  • $\begingroup$ Did u try to look at $Px = \lambda x$ and expand linear equations? $\endgroup$ – Snowball May 11 at 16:06
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$\def\a{\alpha}$ Let $u$ be an eigenvector of the matrix $P$ and assume the elements of the eigenvector have the form: $$ u_k=e^{\alpha k}+ae^{-\alpha k},\tag1 $$ with some parameters $a$ and $\alpha$, which are to be found.

Obviously for all $k=2\dots(n-1)$ $$ (Pu)_k=\frac14\left(e^{\a k}+ae^{-\alpha k}\right)\left(e^\alpha+2+e^{-\alpha}\right) =\frac14\left(e^\alpha+2+e^{-\alpha}\right)u_k.\tag2$$

Thus it remains only to find such $a$ and $\alpha$ that the equation $(2)$ is satisfied for $k=1$ and $k=n$ as well.

For $k=1$: $$\begin{align} &\frac12(e^{\a}+ae^{-\a}+e^{2\a}+ae^{-2\a})=\frac14\left(e^\a+2+e^{-\a}\right)\left(e^\alpha+a e^{-\alpha}\right)\\ &\iff e^{2\a}+ae^{-2\a}=1+a \iff a=e^{2\a}.\tag{3} \end{align}$$

For $k=n$: $$\begin{align} &\frac12(e^{\a(n-1)}+ae^{-\a(n-1)}+e^{\a n}+ae^{-\a n}) =\frac14\left(e^\a+2+e^{-\a}\right)\left(e^{\a n}+a e^{-\a n}\right)\\ &\iff e^{\a (n-1)}+ae^{-\a(n-1)}=e^{\a (n+1)}+ae^{-\a(n+1)}\\ &\iff e^{\a (n-1)}+ae^{-\a (n-1)}=e^{2\a}e^{\a(n-1)}+ae^{-2\a}e^{-\a(n-1)}\\ &\stackrel{(3)}\iff \left(e^{\a (n-1)}-e^{-\a (n-1)}\right)\left(1-e^{2\a}\right)=0\\ &\iff \a_m=\frac{\pi\,i}{n-1}m,\quad m=0\dots n-1 .\tag{4} \end{align}$$

The corresponding eigenvalues are: $$ \lambda_m=\frac14\left(e^{\a_m}+2+e^{-\a_m}\right)=\left(\frac{e^{\frac{\a_m}2}+e^{-\frac{\a_m}2}}2\right)^2=\cos^2\frac{\pi m}{2(n-1)},\quad m=0\dots n-1.\tag5 $$ Since all $n$ eigenvalues are distinct we are done.

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  • $\begingroup$ Thanks, I'm looking at $(5)$ now, how do you have $\frac{e^{\alpha_m/2}+e^{-\alpha_m/2}}{2}=\cos\left( \frac{\alpha_m}{2} \right)$? $\endgroup$ – manifolded May 11 at 20:24
  • $\begingroup$ @manifolded I used: $\frac{e^{ix}+e^{-ix}}2=\cos x$ $\endgroup$ – user May 11 at 20:28
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    $\begingroup$ I see, I missed the $i$ in $\alpha_m$ earlier. Thanks. $\endgroup$ – manifolded May 11 at 20:29
  • $\begingroup$ I see that $(Pu)_k = \frac{1}{4}(e^{(k-1)\alpha}+ae^{-(k-1)\alpha})+\frac{1}{2}(e^{k\alpha}+ae^{-k\alpha})+\frac{1}{4}(e^{(k+1)\alpha}+ae^{-(k+1)\alpha})$, for $k=2,\cdots,n-1$, by definition of matrix multiplication but that is not equal to $(2)$? $\endgroup$ – manifolded May 11 at 20:37
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    $\begingroup$ @manifolded It is equal, with no doubt. $\endgroup$ – user May 11 at 20:40
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(Not a solution)

I numerically found that the answer should be $$ \lambda_j = \sin^{2}\left(\frac{j\pi}{2(n-1)}\right), \quad 0\leq j \leq n-1 $$ but I only have a vague idea to prove this. If we define $B_n = 2P_n - I_n$, then it is enough to show that the eigenvalues of $B_n$ are $$ \cos\left(\frac{j\pi}{n-1}\right),\quad 0\leq j\leq n-1. $$ It seems that the characteristic polynomial of $B_n$ is $$ \phi_n(x) = \frac{1}{2^{n}}(x^2-1)U_{n-2}(x) $$ where $U_n(x)$ is $n$-th Chebyshev polynomial of second kind. However, I'm not sure how to prove this, although it seems that using induction might give recurrence formula for $\phi_n$ which resembles that of $U_n$ a lot.

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  • $\begingroup$ For the numerical solution, may I know if you found $\lambda_j^{(n)}$ for different $n$ and then conclude $\lambda_j$ is what you have above? Thanks. $\endgroup$ – manifolded May 11 at 18:12
  • $\begingroup$ @manifolded Yes, I just plotted the eigenvalues and it reminds me $\sin^2$. $\endgroup$ – Seewoo Lee May 11 at 18:14
  • $\begingroup$ May I know why it is enough for $B_n$ to have its eigenvalues as $\cos \left(\frac{j\pi}{n-1} \right)$? I didn't understand how $B_n=2P_n-I_n$ implies that? $\endgroup$ – manifolded May 11 at 18:17
  • $\begingroup$ @manifolded It follows from $\sin^2(x)=(1-\cos(2x))/2$. $\endgroup$ – Seewoo Lee May 11 at 18:30
  • $\begingroup$ I see, I thought if $B_n$ has eigenvalues of $\cos\left(\frac{j\pi}{n-1} \right)$, then $\frac{I_n-B_n}{2}$ will have eigenvalues $\sin^2 \left(\frac{j\pi}{2(n-1)} \right)$ as opposed to $\frac{I_n+B_n}{2}$ which is what you have. I maybe wrong? $\endgroup$ – manifolded May 11 at 18:39

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